[LintCode] Divide Two Integers 两数相除

最新推荐文章于 2025-02-18 12:22:56 发布

转载最新推荐文章于 2025-02-18 12:22:56 发布 · 69 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：https://yq.aliyun.com/articles/338819

本文提供了一种不使用乘法、除法和取模运算符实现两个整数相除的方法。通过三种不同的C++实现方案，解决了LeetCode上的一道经典题目。这些解法能够有效地处理包括溢出在内的各种情况。

Divide two integers without using multiplication, division and mod operator.

If it is overflow, return 2147483647

Have you met this question in a real interview?

Example

Given dividend = 100 and divisor = 9, return 11.

LeetCode上的原题，请参见我之前的博客Divide Two Integers。

解法一：

class Solution {
public:
    /**
     * @param dividend the dividend
     * @param divisor the divisor
     * @return the result
     */
    int divide(int dividend, int divisor) {
        if (divisor == 0 || (dividend == INT_MIN && divisor == -1)) return INT_MAX;
        long long m = abs((long long)dividend), n = abs((long long)divisor), res = 0;
        int sign = ((dividend < 0) ^ (divisor < 0)) ? -1 : 1;
        if (n == 1) return sign == 1 ? m : -m;
        while (m >= n) {
            long long t = n, p = 1;
            while (m >= (t << 1)) {
                t <<= 1;
                p <<= 1;
            }
            res += p;
            m -= t;
        }
        return sign == 1 ? res : -res;
    }
};

解法二：

class Solution {
public:
    /**
     * @param dividend the dividend
     * @param divisor the divisor
     * @return the result
     */
    int divide(int dividend, int divisor) {
        long long m = abs((long long)dividend), n = abs((long long)divisor), res = 0;
        if (m < n) return 0;    
        while (m >= n) {
            long long t = n, p = 1;
            while (m > (t << 1)) {
                t <<= 1;
                p <<= 1;
            }
            res += p;
            m -= t;
        }
        if ((dividend < 0) ^ (divisor < 0)) res = -res;
        return res > INT_MAX ? INT_MAX : res;
    }
};

解法三：

class Solution {
public:
    /**
     * @param dividend the dividend
     * @param divisor the divisor
     * @return the result
     */
    int divide(int dividend, int divisor) {
        long long m = abs((long long)dividend), n = abs((long long)divisor), res = 0;
        if (m < n) return 0;
        long long t = n, p = 1;
        while (m > (t << 1)) {
            t <<= 1;
            p <<= 1;
        }
        res += p + divide(m - t, n);
        if ((dividend < 0) ^ (divisor < 0)) res = -res;
        return res > INT_MAX ? INT_MAX : res;
    }
};